Phasors
Introduction - What is a Phasor?
Phasors of Time Derivatives
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Introduction

By this time you should know a great deal about DC circuits.  In particular, you should be familiar with the following.

• Resistance
• Ohm's Law
• Series & Parallel Combinations of Resistors
• Kirchhoff's Current Law (KCL)
• Kirchhoff's Voltage Law (KVL)
• The Method of Node Volages
• The Method of Mesh Equations
• Thevinin and Norton Equivalent Circuits
When you learned these topics you might have assumed that what you were learning was something that only applied to DC circuits.  In a way you were right, but in a way you were wrong.  While the methods you learned only apply to DC circuits, the methods you learned can also be generalized to AC circuits.  It will take a little work, but every one of the analysis tools listed above has a related - and very similar - AC version.

If you want to generalize your tool set (all of the tools listed above) you will first need to learn about phasors.  Here is a short lesson on phasors and an introduction to using phasors.

• A sinusoidal signal - voltage or current or any other physical variable - can be represented by a phasor which encodes the amplitude and phase information for the signal.  Call that phasor V.
• If the sinusoidal signal is given by:
• v(t) = Vmax cos(wt + f)
• NOTE:  We could also use:  v(t) = Vmax sin(wt + f).  What matters is the relative phase between signals.
• Then, the signal can be represented by a phasor:
• v(t) <=> Vmax/f
• The little symbol "<=>" means that the phasor - the expression on the right - represents a voltage function of time.
• The voltage function of time is sinusoidal with an amplitude of Vmax, and a phase angle of f.
• and we would write an expression for the phasor V:
• V = Vmax/f
Phasors in linear circuits are related if all of the signals in the circuit are at the same frequency.  If you have signals that are different frequencies you can't use phasors to figure out how signals add, etc.

Important Properties of Phasors - Adding Phasors

In this section we will examine some simple properties of phasors.  We start with one that may or may not be obvious.

• Adding phasors is equivalent to adding the corresponding time function for each phasor.
To see how this works out, consider the sum of two voltage phasors

Vsum= V1 + V2 = V1/f1 + V2/f2

This sum corresponds to the sum of the two voltages (as might occur when you write KVL).  The actual sum would be written:

vsum(t) = Vsum cos(wt + fsum)

= V1 cos(wt + f1) + V2 cos(wt + f2)

The problem is to be able to use the phasor method to do the addition, then interpret that in terms of the time functions.  You and interpret the problem in at least two different ways.
• One way to interpret this is in terms of complex numbers.  The sum voltage - represented as a phasor - is equal to the sum of the other two voltages - represented as phasors.
• Vsum = V1 + V2 = V1/f1 + V2 /f2
• Each phasor can be represented by a complex number.  Break each phasor into real and imaginary parts.
• V1 = V1cos(f1) + jV1sin(f1)
• V2 = V2cos(f2) + jV2sin(f2)
• So, the sum of the two phasors can be computed by adding the real and the imaginary parts separately, giving:
• = V1cos(f1) + V2 cos(f2) + j[V1sin(f1)+ V2 sin(f2)]
• Then, we can note the the real part and the imaginary part are the real and imaginary parts of the sum.
• = Vsum,real + jVsum,imaginary
• And, the phasor for the sum voltage can also be represented with a magnitude-angle representation.
• = Vsumcos(fsum) + jVsumsin(fsum)
There is, of course, also a geometrical interpretation whenever you deal with complex numbers or variables.  If we have two phasors that we are adding, we visualize the situation as shown below.

Vsum = V1 + V2

We can add the two phasors any way possible.  That includes doing it graphically by hand, breaking the phasors into components and summing the real and imaginary components - as we did above - or any other way you can imagine to sum two vector-like quantities.

Differentiation of Time Functions Represented by Phasors

A second operation that we need to perform often is differentiation of a time function represented by a phasor.  Consider a sinusoidal time function, v(t):

v(t) = Vmax cos(wt + f)

with a phasor V:

V = Vmax/f

The derivative of the time function is given by:

vd(t) = dv(t)/dt = - wVmax sin(wt + f)

The phasor for the derivative signal is:

Vd = - wVmax/f-90o= wVmax/f+90o

(Note the minus sign on the first represention where ninety degrees is subtracted has been eliminated when the angle is changed to plus ninety degrees!)

We conclude the following.

• Differentiating a sinusoidal signal is equivalent to multiplication of the signal's phasor by w and rotation of the phasor by 90o.
There is also an interpretation in terms of complex numbers.  Represent the phasor V in terms of complex numbers:

V = Vcos(f) + jVsin(f)

Then, consider the phasor for the derivative:

Vd = Vdcos(f) + jVdsin(f)

But, we can also write the phasor for the derivative from the time function for the derivative.

Vd = wVcos(f+90o) + jwVsin(f+90o)

Now, work with this expression.

• Take the ninety degree terms out of the arguments since:
• cos(f+90o) = -sin(f), and sin(f+90o) = cos(f).

Vd = - wVsin(f) + jwVcos(f)

and with some insight, we can note that this is the same as (since j2 = -1):

Vd = j2 wVsin(f) + jwVcos(f)

Vd = j w( jVsin(f) + Vcos(f)) = j wV

The conclusion:
• Differentiating a sinusoidal signal is equivalent to multiplication of the signal's phasor by jw.
There is a little side note here that is important.  When you study Laplace transforms you will find that differentiating a signal - any signal, not just a sinusoid - is equivalent to multiplication of the signal's Laplace transform by the Laplace transform variable, s.  The similarity is not an accident.  The Laplace transform is a generalization of the Fourier transform, and the Fourier transform is based on the idea that a signal - almost any signal - can be represented as a sum of sinusoids.  In the case of the Fourier transform, that sum becomes an integral.  When you travel this road, you go through Fourier Series, then get to Fourier transforms and finally you get to Laplace transforms.  Not everybody takes all of the steps, and some might jump directly to Laplace transforms, but viewing the entire path can be enlightening.

Example

E1
In a capacitor, the voltage and current are related by:

ic(t) = C dvc(t)/dt

so the voltage phasor, Vc, and the current phasor, Ic, are related by:

Ic = jwCVc

E2   Imagine that you have two voltage sources in series as shown below.

These two voltages are both sinusoidal, but at different phases.  You can see the voltages in this simulator.

• To see the cosine term, type a "1" in the cosine term text-box and be sure that the cosine term stays zero.
• To see the sine term, type a "1" in the sine term text-box, and set the cosine term to 0.